Blog: `3/(x^2 - 3x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Wednesday, May 4, 2016

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$

To decompose a fraction, factor the denominator.

$\displaystyle\frac{{3}}{{{x}{\left({x}-{3}\right)}}}$

Then, write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction.

$\displaystyle\frac{{A}}{{x}}$ and $\displaystyle\frac{{B}}{{{x}-{3}}}$

Add these two fractions and set it equal to the given rational expression.

$\displaystyle\frac{{3}}{{{x}{\left({x}-{3}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}-{3}}}$

To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.

$\displaystyle{x}{\left({x}-{3}\right)}\cdot\frac{{3}}{{{x}{\left({x}-{3}\right)}}}={\left(\frac{{A}}{{x}}+\frac{{B}}{{{x}-{3}}}\right)}\cdot{x}{\left({x}-{3}\right)}$

$\displaystyle{3}={A}{\left({x}-{3}\right)}+\ldots$

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$

To decompose a fraction, factor the denominator.

$\displaystyle\frac{{3}}{{{x}{\left({x}-{3}\right)}}}$

Then, write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction.

$\displaystyle\frac{{A}}{{x}}$ and $\displaystyle\frac{{B}}{{{x}-{3}}}$

Add these two fractions and set it equal to the given rational expression.

$\displaystyle\frac{{3}}{{{x}{\left({x}-{3}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}-{3}}}$

To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.

$\displaystyle{x}{\left({x}-{3}\right)}\cdot\frac{{3}}{{{x}{\left({x}-{3}\right)}}}={\left(\frac{{A}}{{x}}+\frac{{B}}{{{x}-{3}}}\right)}\cdot{x}{\left({x}-{3}\right)}$

$\displaystyle{3}={A}{\left({x}-{3}\right)}+{B}{x}$

Then, plug-in the roots of each factor.

For the factor (x-3), its root is x=3.

$\displaystyle{3}={A}{\left({3}-{3}\right)}+{B}{\left({3}\right)}$

$\displaystyle{3}={3}{B}$

$\displaystyle\frac{{3}}{{3}}=\frac{{{3}{B}}}{{3}}$

$\displaystyle{1}={B}$

For the factor x, its root is x=0.

$\displaystyle{3}={A}{\left({0}-{3}\right)}+{B}{\left({0}\right)}$

$\displaystyle{3}=-{3}{A}$

$\displaystyle\frac{{3}}{{-{3}}}=\frac{{-{3}{A}}}{{-{3}}}$

$\displaystyle-{1}={A}$

So the given rational expression decomposes to:

$\displaystyle-\frac{{1}}{{x}}+\frac{{1}}{{{x}-{3}}}$

This can be re-written as:

$\displaystyle\frac{{1}}{{{x}-{3}}}-\frac{{1}}{{x}}$

To check, express the two fractions with same denominators.

$\displaystyle\frac{{1}}{{{x}-{3}}}-\frac{{1}}{{x}}=\frac{{1}}{{{x}-{3}}}\cdot\frac{{x}}{{x}}-\frac{{1}}{{x}}\cdot\frac{{{x}-{3}}}{{{x}-{3}}}=\frac{{x}}{{{x}{\left({x}-{3}\right)}}}-\frac{{{x}-{3}}}{{{x}{\left({x}-{3}\right)}}}$

Now that they have same denominators, proceed to subtract them.

$\displaystyle=\frac{{{x}-{\left({x}-{3}\right)}}}{{{x}{\left({x}-{3}\right)}}}=\frac{{{x}-{x}+{3}}}{{{x}{\left({x}-{3}\right)}}}=\frac{{3}}{{{x}{\left({x}-{3}\right)}}}=\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$

Therefore, $\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}=\frac{{1}}{{{x}-{3}}}-\frac{{1}}{{x}}$ .

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Wednesday, May 4, 2016

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Wednesday, May 4, 2016

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?